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End of the Proof

(    For every function veS such that/^ v, we have^ g vg, and
f* v dv = I* vg dp by   (,   so   that   $* fg dp  J* vg d^ = J* v dv.
Hence, by definition of the upper integral, we have J* fg d\i g J*/^v, Hence
it remains to establish the opposite inequality

(                               |/dvg \*fgdfj..

Let h e/ be such that h^fg. Then it is enough to show that

(                               \*fdv     hdiJL.

The set X  A is the union of a denumerable increasing sequence of
compact sets Hn and a ju-negligible set N, such that g \ Hn is continuous, finite
and >0 for all n. We define a mapping u of X into R as follows: u = h/g
in the union of the sets Hn, and u(x) = + oo in N and in A. In each of the sets
Hn we have ug = h, and by virtue of (

r*            r**

u dv =      w

JHn               jHn

But v(N) = 0 by virtue of (, and v(A) = 0 by virtue of (,
hence, as h ^ 0, (13.5.7)

r*                  p*                   r*             r*

u dv = sup      u dv = sup      /i dju g     /i djU.

J                       n    JHM                   n     jHn                J

Since/^w, we obtain the required inequality (               Q.E.D.

(13.14.2) Let g be a locally ii-integrable function 'which is ^0 on X, /ef S be
the ^-measurable set of points xeX such that g(x) > 0, and let v = g  /i. Le^
/6e a mapping ofX into R. Then the following conditions (with the conventions
of (13.11) for products) are equivalent:

(a)   / is v-measurable ;

(b)    /<ps w \n-measurable;

(c)    y<7 z,y \i-measurable.

We may suppose # to be finite. With the conventions we have made, we
have jfc = (f(psK09s)> and since the two factors on the right-hand side neverp, = \ fg d^